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Journal of Computer Science and Cybernetics, Vol.22, No.3 (2006), 195—208 THE UNSTEADY FLOW AFTER DAM BREAKING* NGUYEN HONG PHONG1 , TRAN GIA LICH2 1 Institute of Mechanics 2 Institute of Mathematics Abstract. The following problems are presented: the unsteady ﬂow on a river system and reservoirs, the discontinuous wave and unsteady ﬂow after the dam breaking, numerical experiments for some test cases and for natural Da river. T´m t˘t. B`i b´o n`y tr` b`y mˆ h` to´n hoc v` thuˆt to´n t´ d`ng chay khˆng d`.ng trˆn o ´ a a a a ınh a o ınh a . a a. a ınh o ’ o u e hˆ thˆ e o . ´ng sˆng v` hˆ ch´.a, s´ng gi´n doan v` d`ng chay khˆng d`.ng sau v˜. dˆp b˘ ng phu.o.ng ph´p o a o ` u o a . a o ’ o u o a a . ` a d˘c tru.ng, ´p dung t´ thu. nghiˆm sˆ cho bˆn b`i to´n kiˆ m tra c´ nghiˆm giai t´ v` c´c tru.`.ng a . a . ınh ’ e . ´ o ´ o a a e’ o e . ’ ıch a a o ho.p v˜. ho`n to`n c˜ng nhu. khˆng ho`n to`n cua dˆp So.n La trˆn hˆ thˆng sˆng D`. . o a a u o a a ’ a . e e o. ´ o a INTRODUCTION There are several algorithms and softwares for calculating the unsteady ﬂow on a river after dam breaking. Some of them allow calculating the unsteady ﬂow after gradual dam breaking, but cannot exactly determine the position of discontinuous front ξ and the height h of the front ξ (see [8 - 11]). Other ones determine the accuracy of the front ξ and the height h, but the unsteady ﬂow is calculated only on one river branch after the instant dam breaking (see [1, 5, 6, 7]). In this paper the algorithm basing on the [5] permits to calculate the unsteady ﬂow on the river system, connecting with reservoirs after instant or gradual dam breaking. 1. MATHEMATICAL MODELLING The equation system describing the unsteady ﬂows is established from the laws of conser- vation (see [1]) and has the following form: Qdt − ωdx = qdxdt, ∂S S Q2 Q|Q| P+ dt − Qdx = gω i − + Rx dxdt, (1.1) ω K2 ∂S S where h h ∂b(x, ζ) P =g (h − ζ)b(x, ζ)dζ, Rx = g (h − ζ) dζ, ∂x 0 0 ∗ This work is supported by the National Basic Research Program in Natural Sciences, Vietnam
196 NGUYEN HONG PHONG, TRAN GIA LICH x - the coordinate along channel, t - time, q - lateral ﬂow, ω - cross-section area, K - conveyance factor, h - the depth, i - bottom slope, b(x, ζ) - width on the distance ζ from the bottom, g - acceleration due to gravity, S - consideration region, Q - discharge, ∂S - boundary of S. 1.1. One dimensional Saint—Venant equation system If the ﬂow is continuous, from (1.1) we get the Saint—Venant equation system ∂Z ∂Q B + = q, ∂t ∂x ∂Q ∂Q ∂Z + 2v + B(c2 − v 2 ) = Φ, (1.2) ∂t ∂x ∂x where ∂ω gωQ|Q| ∂ω ∂Z 2 gωQ|Q| Φ = iB + v2 − = −B v − , ∂x h=const K2 ∂x ∂x K2 Z - level of free surface, v - velocity, B - width of the water surface, c - celerity of small wave propagation. Equation system (1.2) is quasilinear and of hyperbolic type, which can be rewritten in the characteristic form: ∂Q ∂Q ∂Z ∂Z + (v − c) + B(−v − c) + (v − c) = Φ + (−v − c)q, (1.3) ∂t ∂x ∂t ∂x ∂Q ∂Q ∂Z ∂Z + (v + c) + B(−v + c) + (v + c) = Φ + (−v + c)q. (1.4) ∂t ∂x ∂t ∂x For solving the equation system (1.2) or (1.3)—(1.4), it is necessary to give initial conditions at t = 0 : Z(x, 0) = Z 0 (x), Q(x, 0) = Q0 (x) and the boundary conditions, adjoint conditions. a. Boundary conditions For subcritical ﬂow, one boundary condition is needed: - At the upstream boundary: Q(xb , t) = Qb (t). (1.5) - At the downstream boundary: Z(xb , t) = Zb (t) or Q(xb , t) = f (Zb (t)). (1.6) For supercritical ﬂow: - At the upstream boundary, two boundary conditions are needed: Q(xb , t) = Qb (t) and Z(xb , t) = Zb (t). (1.7) - At the downstream boundary: No boundary condition is needed. b. Adjoint conditions at the internal node of river systems for the continuous ﬂow (for example, nodes D, E, F in Fig. 1.1).
THE UNSTEADY FLOW AFTER DAM BREAKING 197 At every internal node it is necessary to give the following adjoint condition (for example, adjoint conditions at D): A αDj QDj , j∈JD D ZDj = ZD , j ∈ JD , (1.8) F where JD is the set of the river branches having common E node D. C −1 if D is left boundary of the river branch j, B αDj = +1 if D is right boundary of the river branch j. Fig. 1.1 c. Adjoint conditions at the common node A of a river and a reservoir for the continuous ﬂow Suppose that the reservoir has volume V depending on the elevation ZH : V = V (ZH ). The adjoint conditions are (see Fig. 1.2) 2 αj Qj + Q3 = 0, ZAj = ZH , j = 1, 2... (1.9) j=1 dV (ZH ) where Q3 = − . dt 1.2. Adjoint condition at the discontinuous front One adjoint condition at the discontinuous front is needed: (see [1, 5, 7]) 1 [P ]. + [v]2 = 0, (1.10) ω where, [f ] = f + − f − , f − is the value f at the left side of ξ , f + is the value f at the right side of ξ . (1) ξ(t+∆t) . Reservoir ∂S A ξ(t) (2) Fig. 1.2 Fig. 1.3 The velocity of the discontinuous front ξ is (see Fig. 1.3) ω− P + − P − ω+ P + − P − Q+ − Q− C∗ = v + + = v− + = + , ω+ ω+ − ω− ω− ω+ − ω− ω − ω− v + + c+ < C∗ , v − − c− < C∗ < v − + c− . (1.11)
198 NGUYEN HONG PHONG, TRAN GIA LICH In the case when the height of discontinuous front is very small ( h 1), the adjoint condition and velocity C∗ are: Q+ − Q− − B + (v − + c+ )(Z + − Z − ) = 0, (1.12) C∗ = v − + c+ ≈ v + + c− . (1.13) 2. THE ALGORITHMS 2.1. Calculation of the one dimensional unsteady ﬂows (see[2, 4, 5, 6]) Equations (1.3) and (1.4) may be rewritten as follows: dQ dZ dx + a1 = b1 , = c1 , (2.1) dt dt dt dQ dZ dx + a2 = b2 , = c2 , (2.2) dt dt dt where a1 = B(−v − c), b1 = Φ + (−v − c)q, c1 = v − c, a2 = B(−v + c), b2 = Φ + (−v + c)q, c2 = v + c, a. Calculation of the values Z0 and Qk+1 at left boundary L0 k+1 0 • Determine the coordination of point A(i) , (i.e. the intersection of a characteristics line dx/dt = v − c and the line t = tk ) at the iterative step (i) (see Fig. 2.1) τ (0) xA(i) = x0 + (c1 )i−1 + (c1 )A(i−1) , (c1 )L∗ = (c1 )A(0) = (c1 )k 0 . L∗ L 2 0 0 t L0* G L2* tk+1 A(i) B(i) tk L0 L2 T(i) L(i) dx dx =v−c =v+c dt dt Fig. 2.1 • Determine the values ZA(i) and QA(i) by the linear interpolation. • Substituting these values into equation (2.1) we get (i) (i) (i) (i) Q0 + a0 Z0 = d0 . (2.3)
THE UNSTEADY FLOW AFTER DAM BREAKING 199 (i) • From the equation (2.3) and boundary conditions (1.5) one deduces Z0 . (i) (i−1) (i−1) The iterative process is stopped if |Z0 − Z0 | < ε|Z0 |, ε 0.01. b. Calculation of the values ZN and Qk+1 at right boundary L2 k+1 N By the analogous argument from the equation (2.2), we have the following equation at the iterative step (i) (see Fig. 2.1) (i) (i) (i) (i) QN + aN ZN = dN . (2.4) k+1 Solving this equation (2.4) and the boundary condition (1.6) ZN = Zb (tk+1 ) or linearized (i) (i) (i) (i) boundary condition QN + αN ZN = βN , where (i−1) (i−1) (i) ∂f (i) (i−1) ∂f (i−1) (i) (i) αN = − ; βN = QN − .ZN , we get QN , ZN . ∂Z N ∂Z N The iterative process is stopped if (i) (i−1) (i−1) (i) (i−1) (i−1) |ZN − ZN | < ε|ZN |, |QN − QN | < ε|QN |. c. Calculation of the values Z k+1 and Qk+1 at the internal node of river system (for example, D on Fig. 11) For each river branch j (j = 1, 2, ..., JD ) having common internal node D , we have one linear equation (i) (i) (i) (i) QDj = aDj ZDj + dDj , (2.5) where (i) (i) (i) (i) aDj = −(a0 )j and dDj = (d0 )j if D is left boundary of the branch j, (i) (i) (i) (i) aDj = −(aN )j and dDj = (dN )j if D is right boundary of the branch j. From adjoint conditions (1.8) and (2.5), we obtain JD (i) − αDj dDj (i) (i) j=1 ZDj = ZD = JD . (2.6) (i) αDj dDj j=1 (i) (i−1) (i−1) The interative process is stopped if |ZD − ZD | < ε|ZD |. d. Calculation of the values Z k+1 and Qk+1 at the common node of a river and a reservoir (for example, node A on the Fig. 1.2) dV (Z) Linearizing the equation Q3 = − , we have dt V (Z k+1 ) − V (Z k ) 1 dV k+1 Qk+1 ≈ − 3 ≈ − V (Z k ) + (Z − Z k ) − V (Z k ) , (2.7) τ τ dZ (i) Q3 ≈ β (i) (Z (i) − Z k ),
200 NGUYEN HONG PHONG, TRAN GIA LICH 1 dV (i−1) dV (k) where β (i) =− + . 2τ dZ dZ From the equation (2.5) for each river branch, adjoint conditions (1.9) and equation (2.7) we get 2 (i) β (i) ZH − k αj dAj (i) j=1 ZH = 2 . (2.8) (i) β (i) + αj aAj j=1 (i) (i−1) (i−1) Iterative process is stopped if |ZH − ZH | < ε|ZH |. The discharge is calculated from (2.5) and (2.7). e. Calculation of Z and Q at interior nodes of river branch (For example, node G on Fig. 2.1) From the equations (2.1) and (2.2), by method of characteristic we get the following equa- tions for determining the values Z and Q at the iterative step (i) . (i) (i) (i) (i) QG + aL ZG = dL , (i) (i) (i) (i) QG + aT ZG = dT . (i) (i) Solving this equation system we obtain ZG and QG . Iterative process is stopped if (i) (i−1) (i−1) (i) (i−1) (i−1) QG − QG < ε QG and ZG − ZG < ε ZG . 2.2. Discontinuous wave on a river Suppose that the dam sitting at the point L1 is totally and instantaneously broken. The computational process includes (see [1, 6, 7, 8]). a. Calculation of Z − , Q− at the moment of dam breaking According to references [3, 5, 6, 7] these values can be calculated by an iterative method using formulas g 1 1 1 2 Vi = v + + (h(i) )2 − (h+ )2 . + − (i) and hs = v1 + 2 gh1 − Vi , 2 h h 4g where v1 = v(L1 − 0.0), h1 = h(L1 − 0.0), h(i) = h(i−1) + 0.01h+ , h(0) = h+ = h(L1 + 0.0). Iterative process is stopped if hs h(i) . b. Determine the position of the discontinuous front ξ ξ k+1 = ξ(tk+1 ) = ξ k + C• τ, k where ω− P + − P − C∗ = v + + . . ω+ ω+ − ω−
THE UNSTEADY FLOW AFTER DAM BREAKING 201 c. Determine the values (Z + )k+1 , (Q+ )k+1 at the right side of the discontinuous front ξ From the Saint—Venant equation system in the characteristic form (2.1), (2.2) one deduces the equations at the iterative step (i) (i) (i) (Q+ )(i) + aL (Z + )(i) = dL , (i) (i) (Q+ )(i) + aT (Z + )(i) = dT . Solving this equation system we get (Z + )(i) and (Q+ )(i) . We take (Z + )k+1 = (Z + )(i) , (Q+ )k+1 = (Q+ )(i) if |(Z + )(i) − (Z + )(i−1) | < ε|(Z + )(i−1) | and |(Q+ )(i) − (Q+ )(i−1) | < ε|(Q+ )(i−1) |. d. Determine the values (Z − )k+1 , (Q− )k+1 at the left side of the discontinuous front ξ From the equation (2.2) it yields (i) (i) (Q− )(i) + aT (Z − )(i) = dT . Linearizing adjoint condition (1.10) one deduces γ (i) (Q− )(i) + µ(i) (Z − )(i) = θ (i) , where γ (i) , µ(i) , θ(i) , are known coeﬃcients. Solving this equation system we obtain (Z − )(i) and (Q− )(i) . If |(Z − )(i) − (Z − )(i−1) | < ε|(Z − )(i−1) |, |(Q− )(i) − (Q− )(i−1) | < ε|(Q− )(i−1) |, we take (Z − )K+1 = (Z − )(i) , (Q− )K+1 = (Q− )(i) . e. The values Z k+1 and Qk+1 at the boundary nodes, internal nodes of river system, common nodes of a river and a reservoir or interior nodes of each river branch are calculated by the method of characteristic as in the point 1. 2.3. Unsteady ﬂow after the dam breaking on river Suppose that the dam breaking is gradual. The condition at dam is the function: Q = f (ZT , ZD ), (2.9) and QT = QD = Q. (2.10) Linearizing the equation (2.9) we get (i) (i) (i) QT = α(i) ZT + β (i) ZD + γ i . (2.11) a. For the supercritical ﬂow Analogously, from the equation (2.1), (2.2) one deduces two following equations at the left side of the dam (i) (i) (i) (i) QT + aT ZT = dT , (2.12)
202 NGUYEN HONG PHONG, TRAN GIA LICH (i) (i) (i) (i) QT + aL ZT = dL . (2.13) (i) (i) (i) (i) Solving the equations (2.10) - (2.13) we obtain the values ZT , ZD , QT , QD . The iterative process is stopped if the error is small enough and we take (i) (i) (i) ZT = ZT , ZD = ZD , Qk+1 = QT . k+1 k+1 T b. For the subcritical ﬂow From the equation (2.2) at the right side and (2.1) at the left side of the dam we have (i) (i) (i) (i) QT + aT ZT = dT , (2.14) (i) (i) (i) (i) QD + aL ZD = dL . (2.15) (i) (i) (i) (i) Solving the equations (2.10), (2.11), (2.14), (2.15) we obtain the values ZT , ZD , QT , QD . The iterative is stopped if the error is small enough. c. The values Z k+1 and Qk+1 at the boundary, internal, common nodes or interior nodes of each river branch are calculated by the same method as in the point 1. 3. NUMERICAL EXPERIMENTS The method of characteristic is applied to solve some test problems and natural Da river system problem (see [8]). 3.1. Test case 1 Channel of 1.5 km long in which every section is rectangular. Its geometry is described in Fig. 3.1 and Fig. 3.2. The bed slop is about 10% with reverse gradients. One can notice the important contracting section at x = 800 m which creates an acceleration of the ﬂow. This test enables to check that these source terms are correctly evaluated, in the case of ﬂat water at rest. 10 30 9 8 20 7 10 6 5 0 4 0 200 400 600 800 1 000 1 200 1 400 1 600 3 -1 0 2 1 -20 0 0 200 400 600 800 1 000 1 200 1 400 1 600 -30 X(m) X(m) Fig. 3.1. Channel geometry - Proﬁle view Fig. 3.2. Channel geometry - Top view The complete description of the geometry is given in the Table 1. * In each conﬁguration the boundary and initial conditions are as follows: - Downstream boundary and initial condition: level imposed equal to 12 m. - Upstream boundary condition: no discharge. - Initial condition: water at rest at the level 12 m.
THE UNSTEADY FLOW AFTER DAM BREAKING 203 Table 1 Cross-sec X(m) Zb (m) B(m) Cross-sec X(m) Zb (m) B(m) 1 0 0 40 16 530 9 45 2 50 0 40 17 550 6 50 3 100 2.5 30 18 565 5.5 45 4 150 5 30 19 575 5.5 40 5 250 5 30 20 600 5 40 6 300 3 30 21 650 4 30 7 350 5 25 22 700 3 40 8 400 5 25 23 750 3 40 9 425 7.5 30 24 800 2.3 5 10 435 8 35 25 820 2 40 11 450 9 35 26 900 1.2 35 12 470 9 40 27 950 0.4 25 13 475 9 40 28 1000 0 40 14 500 9.1 40 29 1500 0 40 15 505 9 45 * The analytical solution is very simple in this test case. - Water at rest: discharge and ﬂow velocity must be equal to zero. - Flat free surface water level stays at the initial level of 12 m. * The numerical solution (see Fig. 3.3): - Discharge ﬂow is 0 m3 /s. - Water surface level is 12 m. 14 12 10 8 6 4 Numerical Analytical 2 0 0 200 400 600 800 1000 1200 1400 1600 X( m) Fig. 3.3. The numerical solution and the analytical solution 3.2. Test case 2 The steady ﬂow over a bump in a rectangular channel with a constant width. According to the boundary and initial condition, the ﬂow may be subcritical, transcritical with a steady shock, supercritical or at rest. * Geometry data: - The channel width B = 1 m.
204 NGUYEN HONG PHONG, TRAN GIA LICH - The channel length L = 25 m. - Bottom Zb equation x < 8 m and x > 12 m: Zf = 0, 8 m < x < 12 m: Zf = 0.2 − 0.05(x − 10)2 . * Transcritical ﬂow without shock: - Downstream: level imposed equal to 0.66 m, no level imposed when the ﬂow becomes supercritical. - Upstream: discharge imposed equal to 1.53 m3 /s. - Analytic and numerical solution (see Fig. 3.4). * Transcritical ﬂow with shock: - Downstream: level imposed equal to 0.33 m. - Upstream: discharge imposed equal to 0.18 m3 /s. - Analytic and numerical solution (see Fig. 3.5). 1. 2 0.45 0.4 1 0.35 0. 8 0.3 0.25 0. 6 0.2 0. 4 0.15 0.1 0. 2 0.05 0 0 0 5 10 15 20 25 30 0 5 10 15 20 25 30 X(m) X(m) Fig. 3.4 Fig. 3.5 * Subcritical ﬂow - Downstream: level imposed equal to 2 m. - Upstream: discharge imposed equal to 4.42 m3 /s. - Analytic and numerical solution (see Fig. 3.6). 2.5 2 1.5 1 Numerical 0.5 Analytical 0 0 5 10 15 20 25 30 X(m) Fig.3.6. The numerical solution and the analytical solution * Initial conditions - Constant level equal to the level imposed downstream. - Discharge equal to zero.
THE UNSTEADY FLOW AFTER DAM BREAKING 205 - Friction term equal to zero. 3.3. Test case 3 Our purpose is to calculate the unsteady ﬂow resulting from an instantaneous dam breaking in a rectangular channel with constant width. * Geometrical data (see Fig. 3.7): - Channel length 2000 m. - Dam position x = 0 m. - Channel width L=1 m. * Physical parameters - No friction. - Boundary conditions. Downstream: level imposed equal to y2 . Upstream: no discharge. * Initial conditions y = y1 = 6 if x < 0. y = y2 = 0m if x > 0. * Analytic and numerical solution (see Fig. 3.8). 7 6 5 Barrage 4 3 y1 u1=0 u2=0 2 y2=0 1 0 -1500 -1000 -500 0 500 1000 1500 x(m) Fig. 3.7 Fig. 3.8 Dam break on dry bed, initial state The numerical solution and the analytical solution at t = 30 s 3.4. Test case 4 Our purpose is to calculate the unsteady ﬂow of an instantaneous dam break on an already wet bed. * Geometrical data (see Fig. 3.9): - Channel length 2000 m. - Dam position x = 0 m. - Channel width L = 1 m. * Physical parameters - No friction. - Boundary conditions. Downstream: level imposed equal to y2 .
206 NGUYEN HONG PHONG, TRAN GIA LICH Upstream: no discharge. * Initial conditions y = y1 = 6 if x < 0. y = y2 = 2m if x > 0. * Analytic and numerical solution (see Fig. 3.10). 7 6 5 Dam 4 3 y1 u1=0 2 y2 u2=0 1 0 -1000 -800 -600 -400 -200 0 200 400 600 800 1000 X (m) Fig. 3.9 Fig. 3.10 Dam break on wet bed, initial state The numerical and the analytical solution at t =72 s 3.5. Instant dam break and discontinuous wave on the Da river (see Fig. 3.11) 350 Reservoirs 300 250 200 Z(m) 150 100 SonLa Hoa Binh L0 Dam Dam L2 50 0 0 100000 200000 300000 400000 500000 600000 X(m) Fig. 3.11a Fig. 3.11b SonLa dam on the Da river, initial state Da river and the reservoirs 350000 330000 220 310000 210 290000 270000 200 Z(m) 250000 190 Z(m) 230000 180 210000 170 190000 160 170000 150 150000 0 1 2 3 0 0 0.5 1 1.5 2 2.5 3 T(h) t(h) Fig.3.12 Fig.3.13 Z(t) at Son La dam Q(t) at Son La dam
THE UNSTEADY FLOW AFTER DAM BREAKING 207 The Son La dam is situated at the distance of 300 km from the upstream boundary L0 . The water level on upstream side of dam is 215 m and on the other one is 116 m. The cross-section areas are constructed according to the information of ﬁeld measurements. The water volume of Main River and some reservoirs at upstream side of SonLa dam is 9.109 m3 . Suppose that the dam is totally and instantly broken. The numerical solution is presented in the Fig. 3.12, Fig. 3.13 and Fig. 3.14a, Fig. 3.14b. 350 350 300 300 250 250 200 200 Z(m) Z(m) 150 150 100 100 50 50 0 0 0 100 200 300 400 500 600 0 100 200 300 400 500 600 X(km) X(Km) Fig. 3.14a Fig. 3.14b Water surface elevation at t = 0.25 h Water surface elevation at t = 0.5 h 3.6. Gradual dam break and the unsteady ﬂow on the Da river (see Fig. 3.11b) 215 250000 210 200000 205 150000 Z(m) Z(m) 200 100000 195 190 50000 185 0 0 1 2 3 4 0 1 2 3 4 T(h) T(h) Fig. 3.15a. Z(t) at Son La dam Fig. 3.15b. Q(t) at Son La dam 350 350 300 300 250 250 200 200 Z(m) Z(m) 150 150 100 100 50 50 0 0 0 100 200 300 400 500 600 0 100 200 300 400 500 600 X(Km) X(Km) Fig. 3.16a Fig. 3.16b Water surface elevation at t = 0.25 h Water surface elevation at t = 0.5 h
208 NGUYEN HONG PHONG, TRAN GIA LICH The data are given as in the problem 5, we suppose that the dam is gradual failure and rectangular breach 135 m × 105 m (width × depth). Maximum of breach size at t = 0.25 h. The numerical solution is presented in the Fig. 3.15. and Fig. 3.16. CONCLUSIONS The algorithm is applied for calculating: - The unsteady ﬂows of some test cases having the analytical solution. - The unsteady ﬂow on the Da river system after instant or gradual dam breaking. The computational results show that: - The algorithm is easy executed and numerical solutions for the test cases have height accuracy in comparison with the analytical solution. - The unsteady ﬂows on Da river system, connecting with reservoir after gradual dam breaking is corresponded with the ones of other algorithms in [9]. REFERENCES [1] O. F. Vasiliev, M. T. Gladyshev, Calculating discontinuous waves in open channels, Izv. Akad. Nauk, USSR, Mekh. Zh. i G. (6) (1966) 184—189 (Russian). [2] N. E. Khoskin, Methods of characteristics for solving the one-dimensional unsteady ﬂows, Computational Methods in the Hydrodynamics, Moscow, Mir, 1967 (264—291) (Russian). [3] G. Benoist, Code simpliﬁ´ de cacul des ondes de submersion, Rapt E 43/78/55. Lab . e Nat. d’Hydrolique, EDF, France, 1978. [4] A. Daubert et al., Quelques applications de mod`les math´matiques ` des l’´tude des e e a e ´coulements non permanent dans un r´seau ramiﬁ´ de rivi`res ou de canaux, Houille e e e e blanche (7) (1967) 735—746. [5] Ngo Van Luoc, Hoang Quoc On, Tran Gia Lich, Calculation of the discontinuous waves by the method of characteristics with ﬁxed grid points, ZVM and MF, USSR 24 (3) (1984) 442—447. [6] Hoang Quoc On, Tran Gia Lich, Calcul de l’´coulement en rivi`re apr`s la rupture du e e e barrage par la m´thode des diﬀerences ﬁnies associ´e avec des caracteristiques, Houille e e Blanche (6) (1990) 433—439. [7] Tran Gia Lich, Le Kim Luat, Calculation of the discontinuous waves by diﬀerence method with variable grid points, Adv. Water Resource 14 (1) (1991) 10—14. [8] CADAM project (EU Concerted Action on Dam Break Modelling) testcase, 2000. [9] DAMBRK and FLDWAVE are developed by the National Weather Service (NWS), 1998. [10] HEC-RAS is developed by Hydrologic Engineering Center (HEC), US Army Corps of Engineers, 1997. [11] MIKE 11 is developed by Danish Hydraulic Institute (DHI), 2000. Received on October, 2005 Revised on January 20, 2006